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3 years ago

9

Hans's earnings vary directly with the number of hours he works. Suppose that he worked 7 hours yesterday and earned $112. If he

earned $160 today, how many hours did he work today?

1 answer:

densk [106]3 years ago

3 0

Answer: If he earned $160 today the Han worked for 10 hours today

Step-by-step explanation:

112/7=$16= 1 hour of work

160/16=10 hours

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Pamela is 5 years older than Jiri. The sum of their ages is 49 . What is Jiri's age?

tankabanditka [31]

Answer:

Jiri is 22 (and Pamela is 27)

Step-by-step explanation:

Let x = Jiri's age

Then x + 5 = Pamela's age.

If we add the ages together, they said it's 49.

Jiri + Pam = 49

x + x + 5 = 49

combine like terms.

2x + 5 = 49

subtract 5.

2x = 44

divide by 2.

x = 22

So Jiri is 22 and Pamela is 27.

Check:

Pam is 5 yrs older. Check!

22 + 27 = 49 check!

4 0

2 years ago

7. Use the Division Property of Equality to complete the following statement. (1 point)

djyliett [7]

For this case we have the following equation:

$5x = 2y$

To find the value of "x", we use the division equality property, that is, we divide by 5 on both sides of the equation:

$\frac {5x} {5} = \frac {2y} {5}\\x = \frac {2} {5} y$

Thus, using the mentioned property we have to: $x = \frac {2} {5}y$

Answer:

$x = \frac {2} {5} y$

8 0

3 years ago

Find the area bounded by the given curves: <br> y=2x−x2,y=2x−4

Andrej [43]

Answer:

$A = [\frac{32}{3}]$

Step-by-step explanation:

Given

$y_1 = 2x - x^2$

$y_2 = 2x - 4$

Required

Determine the area bounded by the curves

First, we need to determine their points of intersection

$2x - x^2 = 2x - 4$

Subtract 2x from both sides

$-x^2 = -4$

Multiply through by -1

$x^2 = 4$

Take square root of both sides

$x = 2$ or $x = -2$

This Area is then calculated as thus

$A = \int\limits^a_b {[y_1 - y_2]} \, dx$

<em>Where a = 2 and b = -2</em>

Substitute values for $y_1$ and $y_2$

$A = \int\limits^a_b {(2x - x^2) - (2x - 4)} \, dx$

Open Brackets

$A = \int\limits^a_b {2x - x^2 - 2x + 4} \, dx$

Collect Like Terms

$A = \int\limits^a_b {2x - 2x- x^2 + 4} \, dx$

$A = \int\limits^a_b {- x^2 + 4} \, dx$

Integrate

$A = [-\frac{x^{3}}{3} +4x](2,-2)$

$A = [-\frac{2^{3}}{3} +4(2)] - [-\frac{-2^{3}}{3} +4(-2)]$

$A = [-\frac{8}{3} +8] - [-\frac{-8}{3} -8]$

$A = [\frac{-8+ 24}{3}] - [\frac{8}{3} -8]$

$A = [\frac{-8+ 24}{3}] - [\frac{8-24}{3}]$

$A = [\frac{16}{3}] - [\frac{-16}{3}]$

$A = [\frac{16}{3}] + [\frac{16}{3}]$

$A = [\frac{16 + 16}{3}]$

$A = [\frac{32}{3}]$

Hence, the Area is:

$A = [\frac{32}{3}]$

7 0

4 years ago

In every bar graph , line graph , or scatter plot , the vertical axis should include the following

alexgriva [62]

It should include zero

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4 years ago

In Sanford, the high temperature was 54°F on Saturday and 49°F on Sunday. What is the percent of change in the high temperature

ipn [44]

90.74

The percent of change Is the 9%

6 0

3 years ago